nsson until lomorrow at 8 AM CST is allowed , with a one time 10 % penalty to su

Question

nsson until lomorrow at 8 AM CST is allowed , with a one time 10 % penalty to submitted score. Use the References to access important values if needed for this question. For the following reaction, 18.7 grams of nitrogen monoxide are allowed to react with 5.30 grams of hydrogen gas nitrogen monoxide(g) + hydrogen(g)- nitrogen(g) + water(1) What is the maximum amount of nitrogen gas that can be formed? What is the FORMULA for the limiting reagent? grams What amount of the excess reagent remains after the reaction is complete? grams Submit Answer 2. Limiting Reactants: Comparison of Moles Method: This is group attempt 1 of 5 Autosaved at 11:04 AM Back 8

Explanation / Answer

The well balanced reaction is given as :

2NO + 2H2 = N2 + 2H2O

18.7 grams of nitrogen monoxide = 18.7 / molar mass

= 18.7 / 30.0061

= 0.623 mol

5.30 grams of hydrogen = 5.30 / 2.016

= 2.63 mol

Since equimolar amounts of both the reactants are required , so NO is the limiting reagent.

0.623 mol of NO makes 0.623 / 2 = 0.3115 mol of N2

So mass of N2 formed = 0.3116 x 28.0134 = 8.73 grams

0.623 mol of H2 will be utilised = 0.623 x 2.016 = 1.256 grams

Mass of excess H2 left = 5.30 - 1.256 = 4.044 grams

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